lgbasallote
  • lgbasallote
Prove that if n is a perfect square, then n +2 is not a perfect square
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lgbasallote
  • lgbasallote
i suppose first step would be to assume n is perfect square
lgbasallote
  • lgbasallote
so n = m^2
lgbasallote
  • lgbasallote
then i suppose i find n^2

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

lgbasallote
  • lgbasallote
i mean n + 2
lgbasallote
  • lgbasallote
n + 2 would be m^2 + 2
lgbasallote
  • lgbasallote
so now... i assume m^2 + 2 is a perfect square...
anonymous
  • anonymous
we have no 2 perfect square numbers with difference 2
lgbasallote
  • lgbasallote
so i let m^2 + 2 = k^2
lgbasallote
  • lgbasallote
@mukushla that's not really a proof.....
shubhamsrg
  • shubhamsrg
you may continue @igba
lgbasallote
  • lgbasallote
hmm i suppose the next step would be m^2 - k^2 = 2 so (m+k)(m-k) = 2 i don't think this proves anything...
anonymous
  • anonymous
actually that is :) \[m^2-k^2=2\]\[(m-k)(m+k)=2\]and this is impossible
lgbasallote
  • lgbasallote
why so?
shubhamsrg
  • shubhamsrg
well,,you've almost proved it.. now m and k both are integers/.
shubhamsrg
  • shubhamsrg
and its k^2 -m^2 = 2 and not otherwise..
lgbasallote
  • lgbasallote
oh yes
lgbasallote
  • lgbasallote
2 = (k + m)(k-m)
lgbasallote
  • lgbasallote
now...integers huh...
anonymous
  • anonymous
\[m+k>m-k\]so\[m+k=2\]so\[m=k=1\]but it gives\[m-k=0\]
shubhamsrg
  • shubhamsrg
you know k>m.. so from (k+m)(k-m)= 1*2 =>k+m =2 and k-m = 1 this doesnt have an integral solution..
lgbasallote
  • lgbasallote
why do i know k > m again?
shubhamsrg
  • shubhamsrg
m^2 + 2 = k^2 both m and k are positive integers hence k>m
shubhamsrg
  • shubhamsrg
in layman language,, k^2 - m^2 >0
lgbasallote
  • lgbasallote
when was it assumed that m and k are positive?
shubhamsrg
  • shubhamsrg
what else do you mean by perfect square ?
lgbasallote
  • lgbasallote
i don't see how that relates.... the definition of a perfect square is x = k^2
shubhamsrg
  • shubhamsrg
4 is a perfect sq since its 2^2 i.e. square of a positive integer.. and likewise..
anonymous
  • anonymous
@mukushla "we have no 2 perfect square numbers with difference 2" Because 0,1,4,9,6,5,6,9,4,1 etc....
lgbasallote
  • lgbasallote
(-2)^2 is also 4...
shubhamsrg
  • shubhamsrg
you take the absolute value
shubhamsrg
  • shubhamsrg
i mean you have to..
lgbasallote
  • lgbasallote
why?
shubhamsrg
  • shubhamsrg
see,, (k+m)(k-m) =2 either both are positive are both are negative,,since RHS is postive.. so even if you take -ve values,, negative * negative = positive ,,i.e. treated as absolute value..
lgbasallote
  • lgbasallote
...i don't think this is a valid proof....
shubhamsrg
  • shubhamsrg
another explanation might me x is a perfect square because its y^2 x = y^2 = (-y)^2 sqrt(x) = | y | = | -y |
jhonyy9
  • jhonyy9
who can writing here now two perfect squares with difference of 2 ? so because i think do not exist or .... ???
lgbasallote
  • lgbasallote
since when did sqrt x become |-y|
anonymous
  • anonymous
0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats. No +2 in there.
jhonyy9
  • jhonyy9
so and i think that this difference of two will be the ,,key" of this proof of ...
jhonyy9
  • jhonyy9
@lgbasallote what is your opinion from this ,please ?
lgbasallote
  • lgbasallote
my opinion is that this is some tricky algebra
jhonyy9
  • jhonyy9
how do you think it ???
jhonyy9
  • jhonyy9
so and why ?
lgbasallote
  • lgbasallote
because i can't think of a proof
lgbasallote
  • lgbasallote
how can you prove/disprove that the difference of two squares is 2...
lgbasallote
  • lgbasallote
hmmm...wait...
jhonyy9
  • jhonyy9
so i think this is very easy logicaly but to prove it will be difficile i think so but there are again @saifoo.khan and @satellite73 probably they will can doing it hope so much
lgbasallote
  • lgbasallote
their fields of specialty aren't exactly in proving
anonymous
  • anonymous
What's wrong with my proof?
lgbasallote
  • lgbasallote
anyway... m^2 + 2 = k^2 k^2 - m^2 = 2 for it to be equal to 2...k^2 and m^2 should either be both even...or both odd so if i assume both are even.. then k and m are also both even so i can rewrite this as (2x)^2 - (2y^2) = 2 4x^2 - 4y^2 = 2 then... 2x^2 - 2y^2 = 1 x^2 - y^2 = 1/2 then since it's x^2 and y^2 are both integers...the difference should also be integer \('\therefore\) contradiction
lgbasallote
  • lgbasallote
and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..
lgbasallote
  • lgbasallote
@estudier i mean
lgbasallote
  • lgbasallote
@mukushla too
anonymous
  • anonymous
So, that doesn't mean it's invalid..
lgbasallote
  • lgbasallote
it does actually.....you can't do infinite substitutions to prove anything....
anonymous
  • anonymous
There is no infinity involved, all integer squares end in the numbers I gave.
lgbasallote
  • lgbasallote
infinite loops involves infinity agree?
anonymous
  • anonymous
No infinite loop either, just logic.
anonymous
  • anonymous
An integer ends in one of digits 0 to 9. The squares must end in etc....
lgbasallote
  • lgbasallote
but we both know that setting values is illegal in proving
anonymous
  • anonymous
I am not setting any values.
lgbasallote
  • lgbasallote
and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever
lgbasallote
  • lgbasallote
you're checking the squares of each number
anonymous
  • anonymous
I am not.
lgbasallote
  • lgbasallote
and though there is a pattern...there is still susbtitution involved
lgbasallote
  • lgbasallote
and then you'll have to substitute forever to verify that the pattern really does not break
lgbasallote
  • lgbasallote
then infinity is involved
anonymous
  • anonymous
It's not a pattern, it's a fact 0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1
lgbasallote
  • lgbasallote
so you do admit that you checked the 0 to 9 squares
anonymous
  • anonymous
Anything else is impossible.
lgbasallote
  • lgbasallote
...that's substitution
lgbasallote
  • lgbasallote
anyway...my point is...substitution isn't really allowed in proving
anonymous
  • anonymous
You will find that this fact is used a lot in many valid proofs in number theory.
lgbasallote
  • lgbasallote
perhaps
anonymous
  • anonymous
OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.
lgbasallote
  • lgbasallote
since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?
anonymous
  • anonymous
I am not using brute force, brute force is something like testing or experimental math. Or exhaustion (which is also a valid proof technique). It's the use of a fact about squares (ALL squares)
lgbasallote
  • lgbasallote
can you prove it? that it's a fact about squares?
anonymous
  • anonymous
Yes, all numbers end in 0 to 9 and so their squares end in .....QED.
lgbasallote
  • lgbasallote
you didn't really demonstrate anything...you just repeated what you said
anonymous
  • anonymous
You asked for proof, I just gave u a proof.
anonymous
  • anonymous
If you deny my proof, provide a counterexample.
lgbasallote
  • lgbasallote
since you try so hard to defend your proof, i'll accept it
anonymous
  • anonymous
That's very gracious, thank you:-)
lgbasallote
  • lgbasallote
do check on my algebra proof though please. i would like to know if that's the way contradiction works
anonymous
  • anonymous
OK, I will look at it now....
anonymous
  • anonymous
Did u do the case when both are odd? (Perhaps a more straightforward way is one you start with n= some k squared then next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)
lgbasallote
  • lgbasallote
why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?
anonymous
  • anonymous
No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)
lgbasallote
  • lgbasallote
oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement
lgbasallote
  • lgbasallote
again i ask... how is k > m
anonymous
  • anonymous
(k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
anonymous
  • anonymous
factoring involves lesser steps...its obvious that k>m \[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see \[k-m=1\]\[k+m=2\]add them\[2k=3\] Contradiction
lgbasallote
  • lgbasallote
again i ask for the nth time @mukushla how is k > m
anonymous
  • anonymous
ok man i just want to answer ur question
anonymous
  • anonymous
"2 can only be an outcome of even and even or odd and odd" is a valid statement Why do you think it might not be valid?
lgbasallote
  • lgbasallote
well because 8 - 4 is not 2 but it's even minus even
lgbasallote
  • lgbasallote
@mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?
anonymous
  • anonymous
OK, so finally, the algebra Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
anonymous
  • anonymous
i thought they are positive :) sorry
lgbasallote
  • lgbasallote
why (k+1)^2 @estudier ?
anonymous
  • anonymous
I already, said, that's the next square....
lgbasallote
  • lgbasallote
next square?
lgbasallote
  • lgbasallote
is this still related to my proof?
anonymous
  • anonymous
? n is k^2, what is the next square?
anonymous
  • anonymous
no problem ... this is better\[|k|^2-|m|^2=2\]\[(|k|-|m|)(|k|+|m|)=2\]so\[|k|-|m|=1\]\[|k|+|m|=2\]because this time im sure that \(|k|>|m|\) ...now add them\[2|k|=3\]Contradiction and we are done
anonymous
  • anonymous
No, this proof Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
lgbasallote
  • lgbasallote
are you allowed to do that? @mukushla
lgbasallote
  • lgbasallote
can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end
anonymous
  • anonymous
why not...for any real number \[a^2=|a|^2\]
anonymous
  • anonymous
Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 k^2 + 2k +1 >= n +3 n +3 > n+2 QED
anonymous
  • anonymous
That's 2 proofs I have given now...
lgbasallote
  • lgbasallote
a few questions in mind @estudier 1) why find the next square? 2) how is k^2 + 2k + 1 > = n + 3 3) what does n + 3 > n+2 prove?
anonymous
  • anonymous
Think about it....:-)
lgbasallote
  • lgbasallote
in the meanwhile... @mukushla why |k| - |m| = 1?
anonymous
  • anonymous
because\[|k|-|m|<|k|+|m|\]
lgbasallote
  • lgbasallote
how does that make |k| - |m| = 1
anonymous
  • anonymous
u have 2 positive integers with ab=2 and a
anonymous
  • anonymous
right?
lgbasallote
  • lgbasallote
who said those are positive?
anonymous
  • anonymous
|k|>|m| right?
lgbasallote
  • lgbasallote
something feels wrong...
lgbasallote
  • lgbasallote
anyway...where did you get 2|k| > 3
anonymous
  • anonymous
ok...think on what i said...because i believe thats the shortest way to prove ur statement
lgbasallote
  • lgbasallote
|2k| = 3 *
anonymous
  • anonymous
adding\[|k|-|m|=1\]and\[|k|+|m|=2\]
anonymous
  • anonymous
Going back to your question about 2, u can just adjust it to difference of 2q...
lgbasallote
  • lgbasallote
im really skeptical about this... especially that |k|^2 - |m|^2 = 2
lgbasallote
  • lgbasallote
difference of 2q?
anonymous
  • anonymous
You were worried about the validity of your statement, change it to 2q instead of just 2.
anonymous
  • anonymous
Or 2p, or whatever letter u like best.
anonymous
  • anonymous
1) why find the next square? Because we want to see that it isn't n+2 2) how is k^2 + 2k + 1 > = n + 3 Substitute n=k^2 3) what does n + 3 > n+2 prove? See 1)
lgbasallote
  • lgbasallote
oh. when i said difference of two numbers equal to 2
lgbasallote
  • lgbasallote
but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?
lgbasallote
  • lgbasallote
and @mukushla how did you get |k| > |m|
anonymous
  • anonymous
No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.
lgbasallote
  • lgbasallote
and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?
anonymous
  • anonymous
Yes, the algebra is completely general...
lgbasallote
  • lgbasallote
i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...
anonymous
  • anonymous
...which is greater than n+3, right?
lgbasallote
  • lgbasallote
it is?
anonymous
  • anonymous
Number theorists don't pay attention to 0
lgbasallote
  • lgbasallote
unless k is 0....
lgbasallote
  • lgbasallote
why not?
anonymous
  • anonymous
Tradition, convention, call it as u like..... So in fact there are 2 definitions for natural number , one with 0 and one without.
anonymous
  • anonymous
Can I try?
lgbasallote
  • lgbasallote
..you give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....
anonymous
  • anonymous
More the merrier.
anonymous
  • anonymous
"because i know nothing about it" But you do now:-)
lgbasallote
  • lgbasallote
it's not like it's limited to the 99s....
lgbasallote
  • lgbasallote
i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not
anonymous
  • anonymous
Number theory is less onerous than real number stuff because it is integer only. But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.
anonymous
  • anonymous
Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2 Now, (m+1)^2-m^2=2m+1 Since, 2m+1>2 as m is a set of natural numbers. Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.
anonymous
  • anonymous
Yes, that's similar to the one I gave...
anonymous
  • anonymous
"number theorists don't consider 0" http://en.wikipedia.org/wiki/Natural_number
anonymous
  • anonymous
http://mathforum.org/library/drmath/view/63510.html Using 0,1,4, etc...
jhonyy9
  • jhonyy9
ok,so who can answering this question ? ,,the line of perfect squares begin with number ... " what is this number ?
jhonyy9
  • jhonyy9
so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than n+2=m^2 +2 suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 -2 so m1,2 = +/- sqrt(x^2 -2)

Looking for something else?

Not the answer you are looking for? Search for more explanations.